First, let\'s analyze what each component of the solution is (strong base/acid, weak base/acid, salt?) HF: This is a weak acid (something you should probably know off the top of your head, but if you look at Ka, it\'s a very small value which indicates that it is weak). NaOH: A strong base (once again, something you should know off the top of your head. Virtually any ionic compound with Na is strong). Since it\'s strong, we can assume that however much NaOH we have will give us an equivalent amount of OH- ions. NaF: This is the salt for HF (as previously mentioned, this is completely soluble in water and adds F- ions, which will affect the equilibrium of HF). Like before, we can assume that however much NaF we have gives us the same amount of F-, since all of it dissolves. Now, let\'s think about the reactions that are happening. First, whenever you have an acid and a base, you have a neutralization reaction (I\'m leaving Na out of all the following reactions since it is a spectator ion): HF(aq) + OH-(aq) ==> F-(aq) + H2O(l) This means that the OH- we\'ve added will neutralize our initial amount of HF into F- and water. Since we have more HF than OH-, all of the OH- will go towards neutralizing the HF. Thus, 2.00M HF and 1.00M NaOH is essentially equivalent to 1.00M HF and 0 NaOH. Now, all we have is HF and F-. Recall the Henderson-Hasselbalch equation, which tells you how to calculate pH based on the ratio of an acid and its salt. In the case of HF, it will be pH = pKa + log([F-]/[HF]) You have [F-] in the form of [NaF], and you have [HF] which we just determined. You can calculate pKa from the given Ka value. Thus, you should be able to calculate pH. Solution First, let\'s analyze what each component of the solution is (strong base/acid, weak base/acid, salt?) HF: This is a weak acid (something you should probably know off the top of your head, but if you look at Ka, it\'s a very small value which indicates that it is weak). NaOH: A strong base (once again, something you should know off the top of your head. Virtually any ionic compound with Na is strong). Since it\'s strong, we can assume that however much NaOH we have will give us an equivalent amount of OH- ions. NaF: This is the salt for HF (as previously mentioned, this is completely soluble in water and adds F- ions, which will affect the equilibrium of HF). Like before, we can assume that however much NaF we have gives us the same amount of F-, since all of it dissolves. Now, let\'s think about the reactions that are happening. First, whenever you have an acid and a base, you have a neutralization reaction (I\'m leaving Na out of all the following reactions since it is a spectator ion): HF(aq) + OH-(aq) ==> F-(aq) + H2O(l) This means that the OH- we\'ve added will neutralize our initial amount of HF into F- and water. Since we have more HF than OH-, all of the OH- will go towards neutralizing the HF. Thus, 2.00M HF and 1.00M NaOH is essentially equivalent to 1.00M HF and 0 NaOH. Now,.

1. First, let's analyze what each component of the solution is (strong base/acid, weak
base/acid, salt?) HF: This is a weak acid (something you should probably know off the top of
your head, but if you look at Ka, it's a very small value which indicates that it is weak). NaOH:
A strong base (once again, something you should know off the top of your head. Virtually any
ionic compound with Na is strong). Since it's strong, we can assume that however much NaOH
we have will give us an equivalent amount of OH- ions. NaF: This is the salt for HF (as
previously mentioned, this is completely soluble in water and adds F- ions, which will affect the
equilibrium of HF). Like before, we can assume that however much NaF we have gives us the
same amount of F-, since all of it dissolves. Now, let's think about the reactions that are
happening. First, whenever you have an acid and a base, you have a neutralization reaction (I'm
leaving Na out of all the following reactions since it is a spectator ion): HF(aq) + OH-(aq) ==>
F-(aq) + H2O(l) This means that the OH- we've added will neutralize our initial amount of HF
into F- and water. Since we have more HF than OH-, all of the OH- will go towards neutralizing
the HF. Thus, 2.00M HF and 1.00M NaOH is essentially equivalent to 1.00M HF and 0 NaOH.
Now, all we have is HF and F-. Recall the Henderson-Hasselbalch equation, which tells you how
to calculate pH based on the ratio of an acid and its salt. In the case of HF, it will be pH = pKa +
log([F-]/[HF]) You have [F-] in the form of [NaF], and you have [HF] which we just
determined. You can calculate pKa from the given Ka value. Thus, you should be able to
calculate pH.
Solution
First, let's analyze what each component of the solution is (strong base/acid, weak
base/acid, salt?) HF: This is a weak acid (something you should probably know off the top of
your head, but if you look at Ka, it's a very small value which indicates that it is weak). NaOH:
A strong base (once again, something you should know off the top of your head. Virtually any
ionic compound with Na is strong). Since it's strong, we can assume that however much NaOH
we have will give us an equivalent amount of OH- ions. NaF: This is the salt for HF (as
previously mentioned, this is completely soluble in water and adds F- ions, which will affect the
equilibrium of HF). Like before, we can assume that however much NaF we have gives us the
same amount of F-, since all of it dissolves. Now, let's think about the reactions that are
happening. First, whenever you have an acid and a base, you have a neutralization reaction (I'm
leaving Na out of all the following reactions since it is a spectator ion): HF(aq) + OH-(aq) ==>
F-(aq) + H2O(l) This means that the OH- we've added will neutralize our initial amount of HF
into F- and water. Since we have more HF than OH-, all of the OH- will go towards neutralizing
the HF. Thus, 2.00M HF and 1.00M NaOH is essentially equivalent to 1.00M HF and 0 NaOH.
Now, all we have is HF and F-. Recall the Henderson-Hasselbalch equation, which tells you how

2. to calculate pH based on the ratio of an acid and its salt. In the case of HF, it will be pH = pKa +
log([F-]/[HF]) You have [F-] in the form of [NaF], and you have [HF] which we just
determined. You can calculate pKa from the given Ka value. Thus, you should be able to
calculate pH.